Python - 根据文件夹的文件名模式从一个目录中的多个文件夹创建多个 zip 文件

Question

我希望标题不要太混乱。请看下面的图片和gif。

假设我有一个这样的目录。现在我想压缩所有匹配“cs-”、“es-”、“fr-”等的文件夹，如下图所示：

编辑：所有文件夹都有多个子文件夹和/或文件。

到目前为止，我设法找到了我想要的所有文件夹：

import zipfile, os, shutil, pprint

from pathlib import Path


userPath1 = Path(input("Please give me a path:",))

# '- set(userPath1.glob("*.*")' is deducted so that only folders (and no files) are stored in the variables
csFileServerFileList = list(set(userPath1.glob('cs-*')) - set(userPath1.glob("*.*")))
esFileServerFileList = list(set(userPath1.glob('es-*')) - set(userPath1.glob("*.*")))
frFileServerFileList = list(set(userPath1.glob('fr-*')) - set(userPath1.glob("*.*")))
itFileServerFileList = list(set(userPath1.glob('it-*')) - set(userPath1.glob("*.*")))
jaFileServerFileList = list(set(userPath1.glob('ja-*')) - set(userPath1.glob("*.*")))
nlFileServerFileList = list(set(userPath1.glob('nl-*')) - set(userPath1.glob("*.*")))
plFileServerFileList = list(set(userPath1.glob('pl-*')) - set(userPath1.glob("*.*")))
ptFileServerFileList = list(set(userPath1.glob('pt-*')) - set(userPath1.glob("*.*")))
ruFileServerFileList = list(set(userPath1.glob('ru-*')) - set(userPath1.glob("*.*")))
trFileServerFileList = list(set(userPath1.glob('tr-*')) - set(userPath1.glob("*.*")))

langFileServerList = [
                        csFileServerFileList, esFileServerFileList, frFileServerFileList, itFileServerFileList,
                        jaFileServerFileList, nlFileServerFileList, plFileServerFileList, ptFileServerFileList,
                        ruFileServerFileList, trFileServerFileList
                        ]

for list in langFileServerList:
    for i in list:
        print(i)

我知道这种方法：

>>> newZip = zipfile.ZipFile('new.zip', 'w')
>>> newZip.write('spam.txt', compress_type=zipfile.ZIP_DEFLATED)
>>> newZip.close()

但它只适用于文件而不适用于文件夹。

非常感谢您的帮助！

Answer 1

通过修改similar question，我们可以轻松地将目录添加到 zip 文件中：

#!/usr/bin/env python
import os
import zipfile

def create_zip(zip_name: str, paths: list):
    # zipf is zipfile handle
    zipf = zipfile.ZipFile(zip_name, 'w', zipfile.ZIP_DEFLATED)
    # Add each file from each path
    for path in paths:
        for root, dirs, files in os.walk(path):
            for file in files:
                zipf.write(os.path.join(root, file))
    zipf.close()


if __name__ == '__main__':
    # Create a zip with both the tmp/ and tmp2/ folders called myZip.zip
    create_zip('myZip.zip', ['tmp/', 'tmp2/'])

因此，使用您想要压缩的文件夹列表，您将能够做到：

# We need the names of our zip files somehow, here's one way
zip_names = ['cd.zip', 'es.zip', 'fr.zip', ...]

for zip_name, paths in zip(zip_names, langFileServerList):
    create_zip(zip_name, paths)

编辑（完整工作示例）：

#!/usr/bin/env python
import os
import zipfile

def create_zip(zip_name: str, paths: list):
    # zipf is zipfile handle
    zipf = zipfile.ZipFile(zip_name, 'w', zipfile.ZIP_DEFLATED)
    # Add each file from each path
    for path in paths:
        for root, dirs, files in os.walk(path):
            for file in files:
                zipf.write(os.path.join(root, file))
    zipf.close()


def get_dir_list():
    # Get list of all relativ directories
    dirs = [name for name in os.listdir(".") if os.path.isdir(name)]
    # Sort by prefix
    sorted_dirs = {}
    for directory in dirs:
        # We can user the prefix as a key to croup directories together
        prefix = directory.split('-')[0]
        # Check if the key already exists in the sorted dictionary
        if prefix in sorted_dirs.keys():
            # Append the directory to the existing list
            sorted_dirs[prefix].append(directory)
        else:
            # Create the key with a list containing the directory
            sorted_dirs[prefix] = [directory]
    # Returning sorted_dirs.items() will return a list of key value pairs
    return sorted_dirs.items()



if __name__ == '__main__':
    dirs = get_dir_list()
    print(dirs)
    for lang, lang_dirs in dirs:
        create_zip(lang + ".zip", lang_dirs)